In trapezoid $ABCD$, sides $\overline{AB}$ and $\overline{CD}$ are parallel, $\angle A = 2\angle D$, and $\angle C = 3\angle B$.  Find $\angle B$.
Since $\overline{AB}\parallel\overline{CD}$, we have $\angle B+ \angle C = 180^\circ$.  Since $\angle C = 3\angle B$, we have $\angle B + 3\angle B = 180^\circ$, so $4\angle B = 180^\circ$, which means $\angle B = 180^\circ/4 = \boxed{45^\circ}$.

[asy]

pair A,B,C,D;

A = (0,0);

B = (1,0);

D = rotate(120)*(0.8,0);

C = intersectionpoint(D--(D + (40,0)), B--(B + (rotate(135)*(1,0))));

draw(A--B--C--D--A);

label("$A$",A,SW);

label("$B$", B,SE);

label("$C$",C,NE);

label("$D$",D,NW);

[/asy]